NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

### JEE Main 2021 (Online) 27th August Evening Shift

Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.
A
4.18 m
B
2.94 m
C
2.45 m
D
7.35 m

## Explanation

H = $${1 \over 2}$$gt2

$${{9.8 \times 2} \over {9.8}}$$ = t2

t = $$\sqrt 2$$ sec

$$\Delta$$t : time interval between drops

h = $${1 \over 2}$$g($$\sqrt 2$$ $$-$$ 2$$\Delta$$t)2

$$\Delta$$t = $${1 \over {\sqrt 2 }}$$

h = $${1 \over 2}$$g$${\left( {\sqrt 2 - {1 \over {\sqrt 2 }}} \right)^2} = {1 \over 2} \times 9.8 \times {1 \over 2} = {{9.8} \over 4} = 2.45$$ m

H $$-$$ h = 9.8 $$-$$ 2.45

= 7.35 m
2

### JEE Main 2021 (Online) 26th August Evening Shift

A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :
A
hyperbola
B
parabola in the direction of motion of plane
C
straight line vertically down the plane
D
parabola in a direction opposite to the motion of plane

## Explanation

Relative velocity of bomb w.r.t. observer in plane = 0.

Bomb will fall down vertically. So, it will move in straight line w.r.t. observer.
3

### JEE Main 2021 (Online) 27th July Morning Shift

A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $${h \over 3}$$ in both the directions.
A
$${{\sqrt 2 - 1} \over {\sqrt 2 + 1}}$$
B
$${1 \over 3}$$
C
$${{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}$$
D
$${{\sqrt 3 - 1} \over {\sqrt 3 + 1}}$$

## Explanation

$$u = \sqrt {2gh}$$

Now,

$$S = {h \over 3}$$

a = $$-$$g

$$S = ut + {1 \over 2}a{t^2}$$

$${h \over 3} = \sqrt {2gh} t + {1 \over 2}( - g){t^2}$$

$${t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0$$

$${t_1},{t_2} = {{\sqrt {2gh} \pm \sqrt {2gh - {{4g} \over 2}{h \over 3}} } \over g}$$

$${{{t_1}} \over {{t_2}}} = {{\sqrt {2gh} - \sqrt {{{4gh} \over 3}} } \over {\sqrt {2gh} + \sqrt {{{4gh} \over 3}} }} = {{\sqrt 3 - \sqrt 2 } \over {\sqrt 3 + \sqrt 2 }}$$
4

### JEE Main 2021 (Online) 25th July Evening Shift

The instantaneous velocity of a particle moving in a straight line is given as $$V = \alpha t + \beta {t^2}$$, where $$\alpha$$ and $$\beta$$ are constants. The distance travelled by the particle between 1s and 2s is :
A
3$$\alpha$$ + 7$$\beta$$
B
$${3 \over 2}\alpha + {7 \over 3}\beta$$
C
$${\alpha \over 2} + {\beta \over 3}$$
D
$${3 \over 2}\alpha + {7 \over 2}\beta$$

## Explanation

$$V = \alpha t + \beta {t^2}$$

$${{ds} \over {dt}} = \alpha t + \beta {t^2}$$

$$\int\limits_{{S_1}}^{{S_2}} {ds = \int\limits_1^2 {(\alpha t + \beta {t^2})dt} }$$

$${S_2} - {S_1} = \left[ {{{\alpha {t^2}} \over 2} + {{\beta {t^3}} \over 3}} \right]_1^2$$

As particle is not changing direction

So distance = displacement

Distance = $$\left[ {{{\alpha [4 - 1]} \over 2} + {{\beta [8 - 1]} \over 3}} \right]$$

$$= {{3\alpha } \over 2} + {{7\beta } \over 3}$$

### Joint Entrance Examination

JEE Main JEE Advanced WB JEE

### Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

NEET

Class 12